Integrand size = 25, antiderivative size = 99 \[ \int x^4 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx=\frac {d^2 \sqrt {d+e x^2}}{5 (-e)^{5/2}}-\frac {2 d \left (d+e x^2\right )^{3/2}}{15 (-e)^{5/2}}+\frac {\left (d+e x^2\right )^{5/2}}{25 (-e)^{5/2}}+\frac {1}{5} x^5 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \]
-2/15*d*(e*x^2+d)^(3/2)/(-e)^(5/2)+1/25*(e*x^2+d)^(5/2)/(-e)^(5/2)+1/5*x^5 *arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))+1/5*d^2*(e*x^2+d)^(1/2)/(-e)^(5/2)
Time = 0.07 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.73 \[ \int x^4 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx=\frac {\sqrt {d+e x^2} \left (8 d^2-4 d e x^2+3 e^2 x^4\right )}{75 (-e)^{5/2}}+\frac {1}{5} x^5 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \]
(Sqrt[d + e*x^2]*(8*d^2 - 4*d*e*x^2 + 3*e^2*x^4))/(75*(-e)^(5/2)) + (x^5*A rcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/5
Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {5674, 243, 53, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^4 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx\) |
\(\Big \downarrow \) 5674 |
\(\displaystyle \frac {1}{5} x^5 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{5} \sqrt {-e} \int \frac {x^5}{\sqrt {e x^2+d}}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{5} x^5 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{10} \sqrt {-e} \int \frac {x^4}{\sqrt {e x^2+d}}dx^2\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{5} x^5 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{10} \sqrt {-e} \int \left (\frac {d^2}{e^2 \sqrt {e x^2+d}}-\frac {2 \sqrt {e x^2+d} d}{e^2}+\frac {\left (e x^2+d\right )^{3/2}}{e^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} x^5 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{10} \sqrt {-e} \left (\frac {2 d^2 \sqrt {d+e x^2}}{e^3}+\frac {2 \left (d+e x^2\right )^{5/2}}{5 e^3}-\frac {4 d \left (d+e x^2\right )^{3/2}}{3 e^3}\right )\) |
-1/10*(Sqrt[-e]*((2*d^2*Sqrt[d + e*x^2])/e^3 - (4*d*(d + e*x^2)^(3/2))/(3* e^3) + (2*(d + e*x^2)^(5/2))/(5*e^3))) + (x^5*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/5
3.1.12.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[ArcTan[((c_.)*(x_))/Sqrt[(a_.) + (b_.)*(x_)^2]]*((d_.)*(x_))^(m_.), x_S ymbol] :> Simp[(d*x)^(m + 1)*(ArcTan[(c*x)/Sqrt[a + b*x^2]]/(d*(m + 1))), x ] - Simp[c/(d*(m + 1)) Int[(d*x)^(m + 1)/Sqrt[a + b*x^2], x], x] /; FreeQ [{a, b, c, d, m}, x] && EqQ[b + c^2, 0] && NeQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(182\) vs. \(2(75)=150\).
Time = 0.03 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.85
method | result | size |
default | \(\frac {x^{5} \arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )}{5}+\frac {\sqrt {-e}\, e \left (\frac {x^{6} \sqrt {e \,x^{2}+d}}{7 e}-\frac {6 d \left (\frac {x^{4} \sqrt {e \,x^{2}+d}}{5 e}-\frac {4 d \left (\frac {x^{2} \sqrt {e \,x^{2}+d}}{3 e}-\frac {2 d \sqrt {e \,x^{2}+d}}{3 e^{2}}\right )}{5 e}\right )}{7 e}\right )}{5 d}-\frac {\sqrt {-e}\, \left (\frac {x^{4} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{7 e}-\frac {4 d \left (\frac {x^{2} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{5 e}-\frac {2 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{15 e^{2}}\right )}{7 e}\right )}{5 d}\) | \(183\) |
parts | \(\frac {x^{5} \arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )}{5}+\frac {\sqrt {-e}\, e \left (\frac {x^{6} \sqrt {e \,x^{2}+d}}{7 e}-\frac {6 d \left (\frac {x^{4} \sqrt {e \,x^{2}+d}}{5 e}-\frac {4 d \left (\frac {x^{2} \sqrt {e \,x^{2}+d}}{3 e}-\frac {2 d \sqrt {e \,x^{2}+d}}{3 e^{2}}\right )}{5 e}\right )}{7 e}\right )}{5 d}-\frac {\sqrt {-e}\, \left (\frac {x^{4} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{7 e}-\frac {4 d \left (\frac {x^{2} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{5 e}-\frac {2 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{15 e^{2}}\right )}{7 e}\right )}{5 d}\) | \(183\) |
1/5*x^5*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))+1/5*(-e)^(1/2)*e/d*(1/7*x^6/e *(e*x^2+d)^(1/2)-6/7*d/e*(1/5*x^4/e*(e*x^2+d)^(1/2)-4/5*d/e*(1/3*x^2/e*(e* x^2+d)^(1/2)-2/3*d/e^2*(e*x^2+d)^(1/2))))-1/5*(-e)^(1/2)/d*(1/7*x^4*(e*x^2 +d)^(3/2)/e-4/7*d/e*(1/5*x^2*(e*x^2+d)^(3/2)/e-2/15*d/e^2*(e*x^2+d)^(3/2)) )
Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.69 \[ \int x^4 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx=\frac {15 \, e^{3} x^{5} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) - {\left (3 \, e^{2} x^{4} - 4 \, d e x^{2} + 8 \, d^{2}\right )} \sqrt {e x^{2} + d} \sqrt {-e}}{75 \, e^{3}} \]
1/75*(15*e^3*x^5*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - (3*e^2*x^4 - 4*d*e*x ^2 + 8*d^2)*sqrt(e*x^2 + d)*sqrt(-e))/e^3
Time = 0.59 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.06 \[ \int x^4 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx=\begin {cases} - \frac {8 d^{2} \sqrt {- e} \sqrt {d + e x^{2}}}{75 e^{3}} + \frac {4 d x^{2} \sqrt {- e} \sqrt {d + e x^{2}}}{75 e^{2}} + \frac {x^{5} \operatorname {atan}{\left (\frac {x \sqrt {- e}}{\sqrt {d + e x^{2}}} \right )}}{5} - \frac {x^{4} \sqrt {- e} \sqrt {d + e x^{2}}}{25 e} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((-8*d**2*sqrt(-e)*sqrt(d + e*x**2)/(75*e**3) + 4*d*x**2*sqrt(-e) *sqrt(d + e*x**2)/(75*e**2) + x**5*atan(x*sqrt(-e)/sqrt(d + e*x**2))/5 - x **4*sqrt(-e)*sqrt(d + e*x**2)/(25*e), Ne(e, 0)), (0, True))
Time = 0.18 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.40 \[ \int x^4 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx=\frac {1}{5} \, x^{5} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) - \frac {{\left (15 \, {\left (e x^{2} + d\right )}^{\frac {7}{2}} - 42 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{2}\right )} \sqrt {-e}}{525 \, d e^{3}} + \frac {{\left (5 \, {\left (e x^{2} + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x^{2} + d} d^{3}\right )} \sqrt {-e}}{175 \, d e^{3}} \]
1/5*x^5*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - 1/525*(15*(e*x^2 + d)^(7/2) - 42*(e*x^2 + d)^(5/2)*d + 35*(e*x^2 + d)^(3/2)*d^2)*sqrt(-e)/(d*e^3) + 1/1 75*(5*(e*x^2 + d)^(7/2) - 21*(e*x^2 + d)^(5/2)*d + 35*(e*x^2 + d)^(3/2)*d^ 2 - 35*sqrt(e*x^2 + d)*d^3)*sqrt(-e)/(d*e^3)
Time = 0.35 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.02 \[ \int x^4 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx=\frac {1}{5} \, x^{5} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) - \frac {\sqrt {-e^{2} x^{2} - d e} d^{2}}{5 \, e^{3}} - \frac {10 \, {\left (-e^{2} x^{2} - d e\right )}^{\frac {3}{2}} d e + 3 \, {\left (e^{2} x^{2} + d e\right )}^{2} \sqrt {-e^{2} x^{2} - d e}}{75 \, e^{5}} \]
1/5*x^5*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - 1/5*sqrt(-e^2*x^2 - d*e)*d^2/ e^3 - 1/75*(10*(-e^2*x^2 - d*e)^(3/2)*d*e + 3*(e^2*x^2 + d*e)^2*sqrt(-e^2* x^2 - d*e))/e^5
Timed out. \[ \int x^4 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right ) \, dx=\int x^4\,\mathrm {atan}\left (\frac {\sqrt {-e}\,x}{\sqrt {e\,x^2+d}}\right ) \,d x \]